Find the Maximum/Minimum Value y=x^2+2x+5

$y = x^{2} + 2 x + 5$

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{2}{2 (1)}$

Remove parentheses.

$x = - \frac{2}{2 (1)}$

Simplify $- \frac{2}{2 (1)}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$x = - \frac{2}{2 \cdot 1}$

Rewrite the expression.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Evaluate $f (- 1)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = {(- 1)}^{2} + 2 (- 1) + 5$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = 1 + 2 (- 1) + 5$

Multiply $2$ by $- 1$ .

$f (- 1) = 1 - 2 + 5$

Simplify by adding and subtracting.

Subtract $2$ from $1$ .

$f (- 1) = - 1 + 5$

Add $- 1$ and $5$ .

$f (- 1) = 4$

The final answer is $4$ .

$4$

Use the $x$ and $y$ values to find where the minimum occurs.

$(- 1, 4)$

Find the Maximum/Minimum Value y=x^2+2x+5

Related Questions

Support LCMGCF.com by disabling your adblocker.